Choose the right answer. Plane A leaves Tulsa at 2:00 p. m., averaging 300 mph and flying in a northerly direction. Plane B leaves Tulsa at 2:30 p. m., averaging 225 mph and flying due east. At 5:00 p. m., how far apart will the planes be?

Step-by-step explanation:

Average Speed = distance traveled in mile / time in hours

Plane speed = 300 mph = distance traveled by A / time

difference in time from 2:00pm to 5:00p.m = 5: p.m - 2: 00 pm = 3 hrs

distance traveled by A = speed × time = 300 × 3 = 900 miles north

Plane B

time difference = 5: 00 pm - 2:30 pm = 2.5 hrs

distance traveled by plane B = 225 × 2.5 = 562.5 miles east  

to find their distance apart we use Pythagoras theorems

distance apart D = √ ( 900² + 562.5²) = 1061.32 miles

tanФ = 900 / 562.5

Ф = tan⁻¹( 900 / 562.5) = 57.99°

answer: my brother can figure it out, imma go get him so i can answer : ) edit: my brother is sleeping, sorry man

step-by-step explanation:

step-by-step explanation:

divide, then simplify your answer.

1. make -15.66 ÷   5.8 a fraction. then divide.

2. simplify.  

3. final answer.